Measure, Integration & Real Analysis, 2021a

18 Chapter 2 Measures
Outer Measure of Closed Bounded Interval
One more good property of outer measure that we should prove is that if a < b,
then the outer measure of the closed interval [a, b] is b − a. Indeed, if ε > 0, then
(a − ε, b + ε), ∅, ∅,...is a sequence of open intervals whose union contains [a, b].
Thus |[a, b]| ≤b − a + 2ε. Because this inequality holds for all ε > 0, we conclude
that
|[a, b]| ≤b − a.
Is the inequality in the other direction obviously true to you? If so, think again,
because a proof of the inequality in the other direction requires that the completeness
of R is used in some form. For example, suppose R was a countable set (which is not
true, as we will soon see, but the uncountability of R is not obvious). Then we would
have |[a, b]| = 0 (by 2.4). Thus something deeper than you might suspect is going
on with the ingredients needed to prove that |[a, b]| ≥b − a.
The following definition will be useful when we prove that |[a, b]| ≥b − a.
2.10 Definition open cover; finite subcover
Suppose A ⊂ R.
• A collection C of open subsets of R is called an open cover of A if A is
contained in the union of all the sets in C.
• An open cover C of A is said to have a finite subcover if A is contained in
the union of some finite list of sets in C.
2.11 Example open covers and finite subcovers
• The collection {(k, k + 2) : k ∈ Z + } is an open cover of [2, 5] because
[2, 5] ⊂ ⋃ ∞
k=1
(k, k + 2). This open cover has a finite subcover because [2, 5] ⊂
(1, 3) ∪ (2, 4) ∪ (3, 5) ∪ (4, 6).
• The collection {(k, k + 2) : k ∈ Z + } is an open cover of [2, ∞) because
[2, ∞) ⊂ ⋃ ∞
k=1
(k, k + 2). This open cover does not have a finite subcover
because there do not exist finitely many sets of the form (k, k + 2) whose union
contains [2, ∞).
• The collection {(0, 2 − 1 k ) : k ∈ Z+ } is an open cover of (1, 2) because
(1, 2) ⊂ ⋃ ∞
k=1
(0, 2 − 1 k
). This open cover does not have a finite subcover
because there do not exist finitely many sets of the form (0, 2 − 1 k
) whose union
contains (1, 2).
The next result will be our major tool in the proof that |[a, b]| ≥b − a. Although
we need only the result as stated, be sure to see Exercise 4 in this section, which
when combined with the next result gives a characterization of the closed bounded
subsets of R. Note that the following proof uses the completeness property of the real
numbers (by asserting that the supremum of a certain nonempty bounded set exists).
Measure, Integration & Real Analysis, by Sheldon Axler